Question

how do you integrate this using substitution ?
∫x (2x-1)^1/2 dx

Answer 1

try integration by parts

Answer 2

\[u = 2x - 1\]\[x = \frac{u + 1}{2}\]\[dx = \frac{1}{2} du\]\[\int\limits x \left( 2x - 1 \right)^{\frac{1}{2}}dx = \int\limits \left( \frac{u + 1}{2} \right)u^{\frac{1}{2}} \frac{1}{2}du\]\[= \frac{1}{4} \int\limits \left( u^{\frac{3}{2}} +u^{\frac{1}{2}} \right)du\]\[= \frac{1}{4}\left( \frac{2}{5}u^{\frac{5}{2}} + \frac{2}{3}u^{\frac{3}{2}} \right) + C\]substitute back to x